Question

A small body $A$ starts sliding off the top of a smooth sphere of radius $R$. Find the angle $\theta$ (shown in figure above) corresponding to the point at which the body breaks off the sphere as well as the break-off velocity of the body.

• A. $\theta =arc\cos \left(\frac{1}{3}\right)\approx {52}^{\circ }$, $v=\sqrt{\frac{2gR}{3}}$
• B. $\theta =arc\cos \left(\frac{2}{3}\right)\approx {48}^{\circ }$, $v=\sqrt{\frac{2gR}{3}}$
• C. $\theta =arc\cos \left(\frac{2}{3}\right)\approx {48}^{\circ }$, $v=\sqrt{\frac{gR}{3}}$
• D. $\theta =arc\cos \left(\frac{1}{3}\right)\approx {52}^{\circ }$, $v=\sqrt{\frac{gR}{3}}$

Answer 1

Answer: (B)

$\theta =arc\cos \left(\frac{2}{3}\right)\approx {48}^{\circ }$, $v=\sqrt{\frac{2gR}{3}}$

Let us depict the forces acting on the body $A$ (which are the force of gravity $m\overrightarrow{g}$ and the normal reaction $\overrightarrow{N}$) and write equation $\overrightarrow{F}=m\overrightarrow{w}$ via projection on the unit vectors $\widehat{u_t}$ and $\widehat{u_n}$ (figure shown below).
From $F_t=m{w}_t$
$mg\sin \theta =m\frac{dv}{dt}$
$=m\frac{vdv}{ds}=m\frac{vdv}{Rd\theta }$
or, $gR\sin \theta d\theta =vdv$
Integrating both sides for obtaining $v\left(\theta \right)$
${\int }_0^{\theta }gR\sin \theta d\theta ={\int }_0^{v}vdv$
or, $v^2=2gR(1-\cos \theta )$ (1)
From $F_n=m{w}_n$
$mg\cos \theta -N=m\frac{v^2}{R}$ (2)
At the moment the body loses contact with the surface, N=0 and therefore the equation (2) becomes
$v^2=gR\cos \theta$
where $v$ and $\theta$ correspond to the moment when the body loses contact with the surface. Solving equations (1) and (3) we obtain
$\cos \theta =\frac{2}{3}$
or, $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$ and $v=\sqrt{\frac{2gR}{3}}$.